The following text is a slightly reformatted version of a post made by Eric quite a while ago. It has been referred to by so many users (it's now the post for bit-manipulations), and has helped so many that I have reposted it here so it is more easily found.

These operators work on bits and not logical values. Take two 8 bit bytes, combine with any of these operators, and you will get another 8 bit byte according to the operator's function. These operators work on the individual bits inside the byte.

A truth table helps to explain each operation. In a truth table, a 1 bit stands for true, and a 0 stands for false.

The OR operation truth table:

0 OR 0 = 0
0 OR 1 = 1
1 OR 0 = 1
1 OR 1 = 1

The AND operation truth table:

0 AND 0 = 0
0 AND 1 = 0
1 AND 0 = 0
1 AND 1 = 1

The XOR operation truth table:

0 XOR 0 = 0
0 XOR 1 = 1
1 XOR 0 = 1
1 XOR 1 = 0

The NOT operator inverts the sense of the bit, so a 1 becomes a 0, and a 0 becomes a 1.

So let's say I have a byte foo that is initialized to 0:

unsigned char foo = 0;

To set bit 0 in foo and then store the result back into foo:

foo = foo | 0x01;

The OR operation is used between the variable that we want to change and a constant which is called a BIT MASK or simply the MASK. The mask is used to identify the bit that we want changed.

Remember that we write the constants in hexadecimal because it's shorter than writing it in binary. It is assumed that the reader knows how to convert back and forth between hex and binary. ;-)

Usually, though the statement is made shorter in real programming practice to take advantage of C's compound assignment:

foo |= 0x01;

This is equivalent to the statement above.

To clear bit 0 in foo requires 2 bit operators:

foo = foo & ~0x01;

This uses the AND operator and the NOT operator. Why do we use the NOT operator? Most programmers find it easier to specify a mask wherein the bit that they are interested in changing, is set. However, this kind of mask can only be used in setting a bit (using the OR operator). To clear a bit, the mask must be inverted and then ANDed with the variable in question. It is up to the reader to do the math to show why this works in clearing the desired bit.

Again, the statement is made shorter with a compound assignment:

foo &= ~0x01;

To see if a bit is set or clear just requires the AND operator, but with no assignment. To see if bit 7 is set in the variable foo:

if(foo & 0x80)
{
}

The condition will be zero if the bit is clear, and the condition will be non-zero if the bit is set. NOTE! The condition will be NON-ZERO when the bit is set. But the condition will not NECESSARILY BE ONE. It is left to the reader to calculate the value of the condition to understand why this is the case.

There is another useful tool that is not often seen, and that is when you want to flip a bit, but you don't know and you don't care what state the bit is currently in. Then you would use the XOR operator:

foo = foo ^ 0x01;

Or the shorter statement:

foo ^= 0x01;

A lot of times the bit mask is built up dynamically in other statements and stored in a variable to be used in the assignment statement:

foo |= bar;

Sometimes, a programmer wants to specify the bit NUMBER that they want to change and not the bit MASK. The bit number always starts at 0 and increases by 1 for each bit. An 8 bit byte has bit numbers 0-7 inclusive. The way to build a bit mask with only a bit number is to LEFT SHIFT a bit by the bit number. To build a bit mask that has bit number 2 set:

(0x01 << 2)

To build a bit mask that has bit number 7 set:

(0x01 << 7)

To build a bit mask that has bit number 0 set:

(0x01 << 0)

Which ends up shifting the constant 0 bytes to the left, leaving it at 0x01.

MACROS

Because there are a number of programmers who don't seem to have a familiarity with bit flipping (because they weren't taught it at school, or they don't need to know it because of working on PCs), most programmers usually write macros for all of these operations. Also, it provides a fast way of understanding what is happening when reading the code, or it provides additional functionality.

If you are using an unsigned long (32 bit) variable foo, and have to change a bit, use the macro LONGBIT which creates un unsigned long mask. Otherwise, using the BIT() macro, the compiler will truncate the value to 16-bits.[/]

Make Atmel Studio better with my free extensions. Open source and feedback welcome!

You might want to explain what this means. I didn't realize you could actually left shift a bit to a specific place using this actual name of the bit. It drove me crazy because the timer tutorial was directing me here and your tutorial makes no mention of this.

That means, that nr 1 is shifted left as many times as it is needed to reach bit named CS10. Where does the compiler know that CS10 is that? well you give him the AVR's name and it's smart enough to know such things thanks to some smart programmers on the GCC side ;)

For those C compilers that don't have the bit definitions in their .h collection the AVR Studio file \Program Files\Atmel\AVR Tools\AvrStudio4\xmlconvert.exe will probably prove useful to generate .h files from Atmel's own XML part definition files:

Thanks all; I'm going to use most of the above for next yr's class. And another lesson you just taught. Show Dons, BenG, and then Bingos to show writing style and how they accomplish the same thing.
I couldn’t write (copy) the curriculum without this site.
Thanks for the help and making the world a better place,
John

Yup in GCC all your programs include and your Makefile will define a MCU=. As a consequence of these two things it will lead to io.h choosing to pull in one of the io????.h files in \winavr\avr\include\avr\

Yup in GCC all your programs include and your Makefile will define a MCU=. As a consequence of these two things it will lead to io.h choosing to pull in one of the io????.h files in \winavr\avr\include\avr\

Cliff

I am guessing Avr Studio is handling the making the Makefile when I compile my program. Oddly enough I don't even think I defined the correct MCU when I first created the project. I defined it as a Mega 168 and only figured out that it was a Mega 48 when my Dragon complained that the parts didn't match up. Is this because the parts are in the same data sheet?

There's several MAJOR difference between 48 and 168. For example one uses RJMPs for its interrupt vectors and one uses JMPs (because the entire memory is no longer reachable with an RJMP). Far more worryingly the 168 has 1K of SRAM while the 48 has 512 bytes. So the RAMEND used to initialise the stack pointer will be different between the two. I'm therefore astonished that you found that code built for a 168 worked in a 48 !?!

There's several MAJOR difference between 48 and 168. For example one uses RJMPs for its interrupt vectors and one uses JMPs (because the entire memory is no longer reachable with an RJMP). Far more worryingly the 168 has 1K of SRAM while the 48 has 512 bytes. So the RAMEND used to initialise the stack pointer will be different between the two. I'm therefore astonished that you found that code built for a 168 worked in a 48 !?!

Well the code I wrote was only five lines long and didn't use interrupts. Though I might want to make sure that the I actually did program for the 168. I don't have access to the computer I wrote the code in at the moment.

Here I use the XOR operator which is ideal for flipping a bit. Using the 1 XOR as opposed to 1 OR and 1 AND also reduced the code size from 443 bytes (your example) to 398 bytes (my example)

The logic behind it is that on the first run the 8th bit of PORTB, which is 0, is XOR'ed with a 1 which results in a 1.0 XOR 1 = 1
On the next loop iteration the 8th bit of PORTB, which is now 1, will be XOR'ed with a 1 which will result in a 0. 1 XOR 1 = 0

First of all thanks for answer.
I will answer why I need to do this : I writing an dmx receiver and I need to read the address of that receiver from a dipswitch pack. Because I was restricted to single side pcb I have inverted the traces to get the routing done.

Tigrezno,
I am been C ing with AVR for a few months. I am a novice. I got an A/D working with interrupts and putting the data to RS232 to RealTerm on a PC working!
Big accomplishment for me, lots of hours.

I have tried a few of the examples on this topic posted here and ran into problems. Mostly with the fact that PORTs need indirection methods to do bit fiddling? I think?
I just tried your "toggle" on a mega128 PORTA Bit 1 and it compiled and the hardware worked! Thanks, I wanted PORT bit setting and flipping, not just variables. This seems to be the best method discussed here yet? I will try the rest of thems soon.
Where is the #Define documentation for AVR C, or is it per standard C practice and I find it in a C book?
In the DOC directory in my AVR C installation there are many many HTML files. They seem to just compare features to standard C, with no examples? Not much help for a beginner?
Am I missing a manual for this AVR C somewhere?
I didn't know you could use # for parameter subsitution.
Thanks, I put your defines in my first AVR learning C project!
Mike

Thanks, this if from the manual Clawson sent the link too! "I wooda never known!"

"This is called token pasting or token concatenation. The `##' preprocessing operator performs token pasting. When a macro is expanded, the two tokens on either side of each `##' operator are combined into a single token, which then replaces the `##' and the two original tokens in the macro expansion."

I'm a beginner in embedded development and for small applications I really enjoy a simple mode to access a port pin directly ( like PORTA.1 or PORTA_1).
I have found this example a while ago and I would like to hear your comments on it :

Somewhere else on the net i found these two macros beeing called obsolete... Any idea why is it so? Is it indeed more justified to acces the bits as shown in the tutorial?

There are three ways to change a single bit using the AVR. If the register in question is in IO space locations 0x00..0x1F then SBI/CBI can be used to set/clear individual bits. Those opcodes occupy 16 bits and execute in 1 cycle. If the register is between 0x20..0x3F then it is no longer reachable using SBI/CBI but it is using IN/OUT but in this case to set a bit it wil require IN/OR/OUT and to clear a bit it will require IN/AND/OUT which is 3 16 bit opcodes and takes 3 cycles. If the register is beyond IO address 0x3F (that is beyond RAM address 0x5F) then the only access is LDS/STS and to set it wil be LDS/OR/STS and to clear it will be LDS/AND/STS. In this case it's five 16 bit opcodes (LDS and STS are 2 each) and takes 5 cycles to execute.

Now when the early versions of the GCC compiler appeared it did not have an optimisation to spot when LDS/OR/STS could be reduced to IN/OR/OUT or even SBI so a macro sbi() was defined to use inline assembler so the programer could force SBI to be used even though the compiler didn't realise it.

In later compiler versions the code generator was improved so that it would always generate the most optimal code (as long as -O0 was not used!) so the need for the programmer to take control and force when SBI should be used was no longer required.

In the latest GCC (and the other AVR C compilers) the totally portable, totally standard PORTB |= (1<<PB5) will always generate just an SBI PORTB,5 as long as PORTB is between 0x00 and 0x1F and the optimiser is enabled.

The joy of using that standard construct is that you don't need the code to rely on some extra .h file that provides sbi()/cbi() so it makes it more portable between copilers and other architectures.

If using GCC then (for the time being) there is still and this now includes:

so, as you can see, this is just supporting "old" code and turning the macro into the "standard" construct that's been explained above anyway.

As the file is deprecated there's a chance it may be withdrawn from future distributions so it's unwise to write code these days to use it because next time you upgrade compiler you may find that the file has gone.

Is this syntax valid:
PORTC = (0<<PC0); eg. driving the PC0 pin low.
or are you only able to do bit shift operations with "1"
PORTC = (1<<PC0)

I have tried this out on ATMEGA8 and it seems to work but many ppl have told me that that it is not wise to use the bit shift operation with 0 as it can lead to bugs. Could someone please clarify this for me?

Hopefully it should have been obvious to you that to achieve what you think:

PORTC = (0<<PC0);

might do (assuming the intention is only to set bit PC0 to zero) you would actually use:

PORTC &= ~(1<<PC0);

Remeber you can only set bits with | and you can only clear them with &. Though you can obviously set/clear all 8 bits in a register in one go with a simple =, but then you cannot affect just single bits.

I'm working on a project with an external 16-bits adc using SPI. The first 5 bits are for settling of the adc and can be discarded, I am however receiving them.
I'm reading the following bytes in 3 chars (where x is either 1 or 0):
0-7
msb|0|0|0|0|0|x|x|x|
8-15
|x|x|x|x|x|x|x|x|
& 16-23
|x|x|x|x|x|0|0|0|lsb

I want to put 0-7 in a long and shift it 5+16 to the left, put 8-15 in a long and shift it 5+8 to the left and add it to 0-7, put 16-23 in a long and shift it 5 to the left and also add it. Then convert it to a double.

so:
|0|0|0|0|0|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|0|0|0|
<<5
|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|0|0|0|0|0|0|0|0|
typecast
|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|
how can I shift all bits to the left in a long?

When I was a boy... I learned to fix "mainframes" for the Navy. Understand that was 35 years ago so my memory of the details is questionable at best but I seem to remember that a shift took X clocks per shift position. For example to shift one position left took one clock, 2 positions took 2 cycles etc.

Is that true in the Atmega processors, or does it shift any number of positions in a single clock?

The ROL/ROR/LSL/LSR instructions are all shown as being 1 cycle but to shift 5 places might take 5 of them and cost 5 cycles. Though an intelligent optimiser will spot that a shift of 4 can probably be achieved with a SWAP, an AND mask and one shift for the cost of 3 cycles